DIAGONALIZATION/JORDAN FORM

Let A = ( Ibd ),

a. b. c. DEIR

del- ( A - XI ) -_ 72- card )x

c- (ad - Sc )-

dotes )=

Use the quadratic formula

X = ¥±Fad2

The number u-her square root iscalled discriminant( a - d )

'

- 4 (ad - Sc )= a'that d ' - 4¥ Gbc

- Z -d

= ( a - d)'

+ Gbc

-

Theorem ②-

(a) if Ca -d)'+45C So ←

fun we two distinct red eigenvalues.

Then A can he dciagonalizealover KR

(b) If ( a - d)"

t 4 be 20then we two distinct complex conjugateeigenvalue . . Then A can be

diagonalized are G .

(c) If ( a - d )'

-14 be = O c-

Hen c) = is i' repeatede.

'

gen value .

I - th 's case,

e. the -

A = ( T o

u x ) or

A is not diagonal . ' really.

REF of co ③

Suppose A has repeated eczema .

I tent A is diagonalize ble .

So A - p ( I g) p - 'for Sue P

.

Then

A = pie ) p- '

= X p z p- '

= > I = ( I ;)Su tf already diagel .

¢We can analyze case ( C )

('

n more detail . Supposedisc = ( a - d)

'

a- 45=0al A is It diagonal , >all

,

Then A does not have ④2 linearly independent eigenvectors

.

Any eigenvector c 's a multiple

of a single eigenvector V,yo

c- ⇒Set N= (A - > I

.

)

= s

)c dWe ha- Null ( N ) e sp - - v

,

lemma NK O-

z

PI N ? IfH. o

O

= o

n)

If we choose#we t a

⑤

multiple af n, ,

then Nr #O

Howe -e - N ( Nr ) = NK-

= o

Thi 's men , Mr c 's an eigen -aah

so Nr = err,

fo - sore

e -40

Replace ✓ by Lev = Vz

aenNIlet P= ( r

, ma ) .

Then

N P = ( Mr.,Nm ) = (o

,

-,)

= cry. ) ( I ! )in

( A - 72 )P= P ( Ed )h⇒=p(o+xp

A f fer e - pending a - I rearranging , ⑧we get

.

There ( Jordan Canon .cl- Fr-- ?Given a hand .

-

a g - nah - zafc,

2×2 red m- Trix A,

then i 's

aninvert . the netrin P

sunt

a. =p ( Y 'a) p - I#

WA -

- C? I'

.

)Diagonalize or put into

Jordan Canal fun .

- ✓atte - simplify

It (A2) = ( x - 1) IoSu D= I is a repented

eigenvalue

So weneed to use T - C. F . ②

first find a nonzero eigenuih

N= A - I = ( i ? )Nunc : )

,

C ' High 19)

⇒ X - y = O

✓ = t ( t ) Choon C- =/

V,I f ! ) is on e.

'

ge -neck

.

W - t re such tht NK -~,

Van. ( '

o ).

( s --In.Cher , ' Ce s )

P -- (nm ) - ( t j )Thu - A =p ) p - '

Check ⑧'

" "÷.

= ( ii. 11%1=1 ? ;D-

-

Application h s > shes fdiff . eg .

¥,

= A ✓,

If A diagrl.i.lyten can solve

If A nut diaguliznll . . Thenuse J .

C. 7 .

A the - a

change f var -allwe p

- '

✓

Get ⑨II. =L : :b

.

w= ( Wiwa ) Then.

dw .If

= A W,c- W,

d.w(I=>€> t

Yoke we = CeeSuh into previous .

and solve.

h set

w,= C

,te> ←

c- ( q - c,)e> t

Then f . .nl ✓ .

-.

There ( Spectre Theurer,

④- re- sea I)tf tf is a red 2×2 symmetric

matrix ( AT -_ A ),

then.

A can be diagonal . -

z -I a-- IR.

rt a. = (ad) evec d - f A

symmetric.

-

- ti 'd )." ,Then the disc - in.int ← squares

is Itai 30And this i 's zero only whena --

d U 5=0.

Then

A- ( g Oa ) d. eh . ;

di - agent .

one- case, apply first hewn frm④

today to .conclude Ai 's

diagonalize -66 on IR.

4ftp.pl.cnt.cn to calculus

. if 2 reefs.

¥)Lef fcx.gl c 's a

"

n .

-

cu"

fl,

if 2 variably ( hi 've mens conf,

-

nun,

& der . In .exist up to 2nd o-les

al are also continuous )

To find naolm.in Sulu- L

3£ -- o, 3dg - o

( et a sul - tu p = ( xo ,yu )called a critical point .

.

Need 2nd der . -hue test . ④

use Hp -

- (3¥.l . II. b)III. I.

'¥1.

By calculus if 's Symmt- is

The 2nd de - - -true test .

Says , he .he- a

c) Local Max at p .

-f the e. jewels

f Hp are both negative

b) Local Min ab p if eigea uh ,

if Hp are both Ros .hu.

c) ( n remain . ycome test

\

'

n Con c ( u g , he .

Ex fcx, y ) = cost cosy

④

what happens at p - CIO) ?

If P c 's g

If I- scarf cosy Cri 'Ll

¥82 - cosxs.az .

} Point22g

= - cosx cosyetc

.

It.- FEI. !

eigenvalues both neg .

So get locntmx .